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Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in a Banach space $X$. The series $\sum_{n \in \mathbb{N}} x_n$ converges in $X$ if $\sum_{n \in \mathbb{N}} \|x_n\| < \infty$.

Is this statement true?

Since $X$ is a Banach space, if we prove the series $\sum_{n \in \mathbb{N}} x_n$ a Cauchy sequence then it converges. That is, for any $m \ge n > N$, there is $\|\sum_{m \in \mathbb{N}} x_m - \sum_{n \in \mathbb{N}} x_n\| = \|\sum_{k = n}^m x_k\| < \epsilon$ for any $\epsilon > 0$. But how can this be derived from the boundedness of $\sum_{n \in \mathbb{N}} \|x_n\|$?

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  • $\begingroup$ Write the Cauchy condition for the positive series $\sum \|x_n\|$. $\endgroup$ – Giuseppe Negro Sep 24 '18 at 18:09
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Triangle inequality: that follows from the fact that$$\left\lVert\sum_{k=n}^mx_k\right\rVert\leqslant\sum_{k=n}^m\lVert x_k\rVert.$$

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  • $\begingroup$ yes but how to show $\sum_{k = n}^m \|x_k\| < \epsilon$ from $\sum_{k = n}^\infty \|x_k\| < \infty$? $\endgroup$ – Analysis Newbie Sep 24 '18 at 20:10
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    $\begingroup$ The sequence $\left(\sum_{k=1}^n\lVert x_k\rVert\right)_{n\in\mathbb N}$ is increasing and bounded. Therefore, it converges. It follows that it is a Cauchy sequence. $\endgroup$ – José Carlos Santos Sep 24 '18 at 20:14
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In Banach spaces, absolute convergence implies convergence. Since $s_{n} = \sum_{k=0}^{n}\lVert x_{k}\rVert$ in monotonic and bounded, it converges. Therefore the given series $\sum_{n\in\mathbb{N}}x_{n}$ converges:

About Banach Spaces And Absolute Convergence Of Series

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